The locus of the centre of the circle described on any focal chord of the parabola `y^(2)=4ax` as the diameter is

Options

A. `y^(2)=2a(x+a)`
B. `y^(2)=a(x+a)`
C. `y^(2)=2a(x-a)`
D. `y^(2)=4a(x-a)`

Answer

To find the locus of the center of the circle described on any focal chord of the parabola \( y^2 = 4ax \) as the diameter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parabola and its Focus**:
The given parabola is \( y^2 = 4ax \). The focus of this parabola is at the point \( (a, 0) \).

**Hint**: Recall that the focus of a parabola \( y^2 = 4px \) is located at \( (p, 0) \).

2. **Consider a Focal Chord**:
A focal chord is a line segment that passes through the focus and has endpoints on the parabola. Let the endpoints of the focal chord be \( (at_1^2, 2at_1) \) and \( (at_2^2, 2at_2) \) where \( t_1 \) and \( t_2 \) are parameters related to the points on the parabola.

**Hint**: The endpoints of the focal chord can be expressed using the parameterization of the parabola.

3. **Find the Midpoint of the Focal Chord**:
The midpoint \( P \) of the focal chord, which will be the center of the circle, has coordinates:
\[ H = \frac{at_1^2 + at_2^2}{2}, \quad K = \frac{2at_1 + 2at_2}{2} = a(t_1 + t_2) \]

**Hint**: Use the midpoint formula to find the coordinates of point \( P \).

4. **Use the Property of Focal Chords**:
For a focal chord, the product of the parameters of the endpoints is \( t_1 t_2 = -1 \). Therefore, we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = -\frac{1}{t_1} \]

5. **Substitute \( t_2 \) into the Midpoint Coordinates**:
Substitute \( t_2 \) back into the expressions for \( H \) and \( K \):
\[ H = \frac{a(t_1^2 - \frac{1}{t_1^2})}{2}, \quad K = a\left(t_1 - \frac{1}{t_1}\right) \]

6. **Simplify the Expressions**:
Rewrite \( H \) and \( K \):
\[ H = \frac{a(t_1^4 - 1)}{2t_1^2}, \quad K = a\left(\frac{t_1^2 - 1}{t_1}\right) \]

7. **Eliminate the Parameter**:
From the expression for \( K \), we can express \( t_1 \) in terms of \( K \):
\[ t_1^2 = \frac{K}{a} + 1 \]
Substitute this back into the expression for \( H \) to eliminate \( t_1 \).

8. **Derive the Locus Equation**:
After substituting and simplifying, we will arrive at the equation of the locus of the center \( P \):
\[ K^2 = 2a(H - a) \]
or equivalently,
\[ y^2 = 2a(x - a) \]

**Hint**: Recognize that the final equation represents a parabola.

### Final Result:
The locus of the center of the circle described on any focal chord of the parabola \( y^2 = 4ax \) as the diameter is given by:
\[ y^2 = 2a(x - a) \]